 Well, let me thank the organizers for inviting me and giving me the chance to visit Italy, which I haven't been in a very long time, and I guess as I'm the last talk, perhaps we can all thank the organizers who have done a tremendous amount of work and done a terrific job. I want to give my personal thanks to both Mel and Craig for many years of guidance, support, and friendship. So thank you, both of you. I also wanted to point out that we're celebrating their hundred and fiftieth birthday, more or less, this year, and also, if you count from their PhDs, that's a hundred years of mathematics. So those are important things. And then, I guess, following up on Long's desire to speak to the young people, my advice, it's not advice, it's just pointing out that it's okay to be nervous before your talk. It's probably should be expected. I've been doing this a long time, I still get a little nervous. All right, so I want to talk on a paper that's joint with Craig and my student, Thomas Poulstra, on the equality of test ideals. And so let me start with a non-characteristic P question. And I'm going to assume, pretty much, that we're in a complete domain, D is the dimension of R. So let's let I contained in R be a height one and unmixed, okay? And let's set, I'll give it a name so I can refer to it B, sub J, and let's let that be the annihilator over R of the jth local cohomology of R mod I symbolic N, okay? So the question is, how can we control this ideal? All right? So what you can see is that local duality says that the the height, the height of these guys is at least, I guess, D minus J plus one. You need the symbolic power there for that. And so, all right, so of course, if you take, if we take one to be in I, all right, then that tells us that X1 to the N is certainly in this ideal B, sub J, N, all right? But the question, first question that I want to ask is, can we find parameters, come up here, there's my page two, parameters X1, X2, up to X sub, we want to go as far as B minus J plus one in the first one, say, BJ1, such that they're nth powers in this ideal BJN, okay? And the issue is that these, the exponents, these exponents, I mean, there's some exponent that will get you in there, but it's, it's hard to know up priority what those exponents are, okay? So that's, that's our, so I want you to kind of tuck that away and now we'll start the tight closure part of the talk, okay? So for the rest of the talk, I want to assume still that I have this complete domain, but now I'm positive characteristic P and F finite and we've seen this definition lots of times, but R is weakly F regular, if and only if I star equals I for all ideals in R and R is F regular if Rp is weakly F regular for all primes P in spec, okay? And as has been mentioned several times before, there's an example of Brenner and Monsky that we have, so there exists R, a ring, W, a multiplicative set, I, an ideal in R, such that we have that if we take the tight closure of I and R and we localize, then that will always be contained in the tight closure of the localization in the ring, in the localized ring, but it's strictly, right, strictly contained. So localization is bad and I asked Holger earlier and he thinks that, so this is an affine example, but he thinks we can, it can be homogenized and made into a local example. So I hadn't been sure about that. All right, so because of that example, this weakly F regular doesn't imply F regular, very straightforwardly at least. So Hoxter and Heunicke just defined what strongly F regular means, so let me remind you what that means. So we have a strongly F regular ring if for all C in R naught, right here for us, we're assuming a domain, so just assume it's not zero, there exists Q sufficiently large, Q is a P to the E power, such that R times C one over Q sitting inside R one over Q, that's the meaning of strongly F regular and this localizes well. You can turn this into a question about hams, everything is finite, nice localization. Okay, and this was the key to finding test elements, essentially, among other things. So we obviously have that, well not obviously, but strongly F regular implies, well implies F regular, which obviously implies then weakly F regular. Okay, and also I just want to point out, this implies normal and Cohen-McCauley. Okay, so I'm interested in the question of completing the circle here, I'll write that down in a second, so let's do that. So question two, I should say does weakly F regular imply strongly F regular? Okay, so let me give you a brief history of this. So in the original paper in 1990, Hoxter and Heuneky show this if R is Gornstein, I mean the paper was published in 1990, I'm sure this was known by them as soon as they decided to talk about strongly F regular. Then the next result was in 1993, Lori Williams, a student of Mills, showed that it's true in dimension, for dimension of R less than or equal to three, well she really did the dimension three case, and in 1995 another student of Mills, Brian McCriman, did the case that R is Q Gornstein on the punctured spectrum, and in fact what Williams showed is that that's actually what Williams showed because that's true always in dimension three, and McCriman generalized that, and then Brian came actually to Missouri and spent some time as a postdoc, and so that was in 1996 or fall of 95 maybe, and so that was when I started working on this problem, so I tell my students in class you know you think you worked on that problem a long time by working on it for an hour, but let me tell you what a long time is, yeah so a little bit more history in 1999, Gennady Lebesnik and Karen Smith did the, but the argument is highly dependent on the grading, it's impossible to generalize it beyond that, and I want to mention also around the same time, I think, did I write down here, so around 2000 Adela Virigiu, another student of Mills, had related results, so related graded results, the same result, and that's where things stood for a long time, there really hasn't been any progress since then that I'm aware of, okay, so let me spend some time trying to explain to you for where, what's going on in this problem, so let me take an injective hull, and so basically tight closure behavior in R is, this might be a bit of the wrong word, but let me say it's determined by the tight closure of zero in E, so don't take that too rigorously, there's no rigor there, okay, I mean for those who know this is the annihilator of this is the big test ideal, that's a very concrete thing, but somehow overall the way this module behaves tells you something about tight closure overall, and so the one issue here is that E is not finitely generated, well outside of dimension zero, and so there are a couple of tight closures we have to think about, so the first one, so let me just say there are several tight closures, alright, so there's what's called the finiteistic tight closure in E, and so what that is, is the union over all say M contained in E and M is finitely generated, we look at the tight closure in M, we take all of those guys, and then there's also the tight closure in E, which I'm going to describe how we can think about this without having to understand anything about tight closure in modules, because nobody really wants to think about general modules in this case, alright, so we do have that the finiteistic tight closure is contained in the real tight closure, the true tight closure, and so of course we really what I think we'd like to see is that these are equal, that's essentially what we're after, okay, I guess I'll post state that as a question soon, and their connection, the direct connection to the questions of weak and strong F regularity are as weakly F regular, if and only if this finiteistic tight closure is zero, that reflects something we can talk about that doesn't involve localization, and are strongly F regular if and only if the actual tight closure is zero, and so question three is is zero star E equal to the finiteistic tight closure, that's the question we'd like to answer, alright, and so yes to question three is going to imply yes to question two, let me go back to question two briefly, I missed question two, where did I put question two, does weak F regularity imply strong F regularity, so alright, so let's assume that our local ring, I guess I've already assumed this, but let me emphasize that RMK is a Cohen-McAuley domain, and it has, and J contained in R is a canonical ideal, essentially I'm assuming we have enough to get a canonical module, and then we can always represent that by an ideal when we're in the domain case, alright, so we can construct the injective hall in the following way, so let's choose, we're going to choose a system of parameters, so say x1 in J, and then x2 through xd, take a system of parameters, alright, and I'm going to need their product, so let's call y1 the product of the x's, and we're going to set this ideal i sub t to be, we're going to take J, multiply it by the t minus first power of x1, and take the t-th powers of the rest of the x2's, and then in this case, each r mod it, well when you have a canonical ideal and you kill it, you get a Gorenstein ring, and x1 to the t minus 1 J is isomorphic to J, so we still get a Gorenstein ring, and then we're killing a system of parameters in the Gorenstein ring, so this each rit is a zero-dimensional Gorenstein ring, that's supposed to say ring, my pen isn't writing there for some reason, so choose u1, let's call it in r, a Sockle generator mod i1, and then let's set ut to be the product, y1 to the t minus 1 u1, and then this is, this guy will be a Sockle generator mod it, so we have this picture, we have r mod i1 and r mod i2 for instance, and here there's u1 bar, finished teaching linear algebra, but it's not a vector, so u1 sits in here, and you can map it to u2, and so there's an injection like that, and this keeps, we can then inject into r3, so u2 goes to u3 bar, and so on, so in here is ut bar, and the the union, the direct limit gives you e, so e is the direct limit here of the r mod i2, and so using this construction I'm going to be able to put everything I want to talk about in tight closure in terms of ideals, which is a lot easier to think about, so the finiteistic tight closure in this case turns out to be the direct limit over t of the it star mod the it, so that's easy to describe, and so let me say what this means, so if we have w plus some it1, if that's in the finiteistic tight closure of e, so this is saying the following statement, there exists a c and r0, and there exists a t2 such that for all q, the following c, y to the t2w to the q is in i to the t1 plus t2q, all right, so this is a direct limit, so being in the tight closure mod it1, you might not be in the tight closure mod it1, but if you push some finite number of steps you'll get in the tight closure, so but versus the following, what does it mean for w plus it1 to be in the tight closure of e? Well, so this says there is a multiplier, there's a c and r0, we get to start off the right way, but now we have to switch quantifiers, so such that for all q there exists, let's call it an sq, and this sq depends on q such that c, y to the sq, w to the q is in i, now we've got t1 plus s, so q, all right, and sq depends on q, and this is where the complications come in, because what we'd really like to be able to do is to take that sq and say, oh no, there is a uniform bound on the sqs that's independent of q, all right, so, skip what I was going to do there for time, all right, so now let's go back to, this is very general what I've been talking about, if one is really interested in the bigger picture, not just of is weak and does weak imply strong, but does the so-called test ideal and the finitely generated test ideal do those, are those the same, all right, so let's go back to the weakly F regular case, so what does it mean that we're not weakly F regular, or that we are weakly F regular, so this says for all say t0, this element u t0 is, which again, right, is y t0 minus 1 u1, that is not in i t0 star, that's the condition of F regular, strong F regularity actually, all you have to do is look at the soccer element and it says that u1, I guess bar, is not in the tight closure of 0, that's it, all right, so let's suppose that u1 bar is in the tight closure, so what does that tell us, that tells us that there exists a c, say not equal to 0, such that for all q there exists, I guess I called it s, such that, I lost my place here, okay, such that all we need to do is we've got c sq u1 raised to the q is in i to the, I guess this would just be 1 plus sq, that's the t, all right, and let me remind you, so this is x1, I guess, to the sq, all right, let me, y1 to the sq with the c u1 to the q, oh, damn it, try again, it's over here, I want to say this is c y1 to the sq times u1, sorry, I think that makes, that actually makes sense now, I can tell my class yell at me if I make a stupid mistake, they never do, and so how am I doing 20 minutes, I think I'm going to make it, so when you, I want to look at this, so this is the ideal x1 to the 1 plus sqj, x2 to the sq plus 1 through xd to the sq plus 1, raised to the q, oh yeah, yeah, thank you, that's very important, that should just be an sq, okay, now, so, all right, so let me call y2 the product from x2 on, all right, so y1 is just x1, y2, all right, and now what you get from this is you can write this, this way, c y2 to the sq u1 to the q is in j to the q plus x2 to the sq plus 1 through xd to the sq plus 1 to the q, and I'm colonning here with x1, I'll just put in x1 infinity, some big power of x1, but this is, these guys wear a colon Macaulay ring and these are parameters, so this is just equal to the ideal, so you see we've taken x1 out of the picture without any trouble, so what we have now is the following picture, I have, think about it this way, I have r mod j x2 through xd to the q and then, since I've, again, since I've taken x1 out of the picture, I have j and I get x2 squared through xd squared to the q and this keeps going, and over here we've got, say, j x2 squared xd, sorry, this should be an sq and an sq raised to the q, and this is all multiplication by y2 to the q each, so what we have after multiplying by c is we have something here that, as we push it forward, eventually here it goes to zero, so this is, this is essentially, we're looking in a direct limit, we've got an element in the direct limit that we know is zero, and therefore, it, well we know already in some sense that sq is the finite value that puts us into zero, but the problem is we would like to not have sq depend on q, we would like to, to bound it, all right, and so let me tell you, I, I, I don't want to kind of get too technical, it, it quickly gets hard to describe, but let me just say that, so we, we can get x, the exponent on x2 down easily, all right, so r is normal, okay, so, so pick x2 such that x2 times j is contained in a principal ideal, okay, which we can always do, and then we can bound the exponent on x2, and then in dimension three, sorry, at height two, so that's a result of Lippmann, that if we're dimension two, and I, I believe it the correct term is pseudo-rational, all right, which holds in this case, f-rational singularities in dimension two are, are very nice, that says that the, the class group is torsion, maybe it says more than that, finite, but all I need is torsion, and that means we can bound the exponent, whoops, the exponent on x3, we can, or we can choose an x3 and bound the exponent, all right, and that's, that's actually, that's essentially what Laurie Williams argument was, so in dimension three we're done, and then if you assume that, that the canonical ideal is torsion at all heights up to d minus one, then essentially you can keep bounding the exponents on the higher, you can choose the x's and keep bounding the exponents, oh, I skipped ahead, all right, so let's, we can erase this, because I just wrote that down, so let me get back in on schedule and my notes, so I won't write much on this page, so the problem is for xi, i bigger than or equal to four, we need a new idea, okay, this, this, this kind of thing isn't going to work, and so let's say that our j is the canonical ideal, as I said, well, you can take what's called, you can find another ideal that intersects with j and gives you a principal ideal, so this is the anti-canonical ideal, I guess this linkage, right, so I brought, I brought the parents work in, so they're a whole theory of linkage in height one, right, so notice that, and I want to do this in such a way that the height of i plus j is two, so they don't share any components, all right, okay, so I guess here is where the, the new kind of stuff starts, I'm a little behind where I wanted to be, so let me see what I can say, so the setup here is let's think of R as the quotient of a regular local ring, all right, and I think back here, let me add in a little bit here, this j bracket q, I can replace this by actually the q-symbolic power, if I can do it in that can the q-symbolic power is bigger, it turns out that's fine, so, so what we do is we consider the following, we stick our mod j to the symbolic power into its double-dual over s, so we look at x h over s of h over s of r mod jq, and this is all into s, s is a nice Gorenstein ring, so, so that's its double-dual, and let's, let's call this mq because I don't want to have to write it again, and so, oh, and I forget, I'm gonna make you go back in your notes again, and this is page 10, call this last thing equation one, okay, I'll leave that there for a minute, all right, so, it turns out that if you want to, the bounding exponents in equation one is essentially the same as doing the same, bounding the exponents, exponents on mq, well, the local co-emology of mq rather than, I mean, it's not on mq itself, and so, this is equivalent by local duality, and using the Cohen-McCauley-ness of the regular ring, and what I call much pain, which, that's why this paper is so hard to read, this is the same thing now as bounding annihilation, local co-emology modules over m of the r-mod iq symbolic powers, and so, we are back to question one, that is why I wanted to know, an answer to question one is that we changed the question from, into this much simpler looking question about local co-emology modules on this one ideal, okay, so, let me tell you, let me spend the last six minutes saying what we were able to prove, so, let me call them, I'll start with theorem one, so suppose we have a local ring that's excellent, and Cohen-McCauley normal, and dimension at least four, take i to be this anti-canonical ideal, and let's assume that it's, the class group is torsion in co-dimension two, so then what we can show is for, between one and d minus two, so there exist parameters x1 through xd minus j plus one, such that we get annihilation in the following, so if we look at the ideal x1 through xd minus j plus one, we raise that to the q-th power, then if we look at that times the j-th local co-emology of, well, we, we have to include that m, it turns out, so we look at i to the mq, then that is equal to zero, for e bigger than or equal to one, oh, well, for all q, and in particular, what we're saying is this finiteistic tight closure over e of e is the regular tight closure, so, so getting these bounds on those, these local co-emology modules of Armand symbolic powers is sufficient to determine, now this is, this turns out to be pretty hard to, to satisfy, or to satisfy the conditions of the theorem, we, we don't know very many cases, but let me tell you one case, so the case that we were able to, to get, so let's assume that we've got an excellent weakly-affirregular ring of dimension d, let's take i to be our anti-canonical ideal, and the, the condition on, this is an assumption in theorem one that we have such a thing, but in a weakly-affirregular ring that, that assumption is satisfied, okay, so, I want to suppose the following, so we look at the, the so-called anti-canonical cover, okay, so what this is is the direct sum of the symbolic powers, so suppose that that is noetherian, but we only need on the punctured spectrum again, okay, so the conclusion is that there exist parameters as in theorem one, i.e., r is strongly-affirregular, okay, so this, this is a new result that does push the envelope epsilon farther, since the, the, well there, there are lots of cases where the, the anti-canonical cover is not noetherian, but I guess the last thing I want to mention is something I admit to not understanding the mathematics very well, but I, the minimal model program I guess does have something to, to say here, so there's a result of Doth's and Waldron that for characteristic p greater than five, if, if the dimension of the ring is three, and if the ring is essentially a finite type over a field of infinite transcendence degree over Zp, then this ring, the anti-canonical cover, this guy, is noetherian, and my, my understanding is the conjecture, no, no, no, sorry, let me move that down, that's, that is much too strong a statement, it's not right, yeah, no I need one assumption, but a very important assumption, yeah, as soon as I started to think about going on, if R is strongly F regular, which is a strong assumption, then this, this anti-canonical cover is noetherian, and I gather the conjecture is they, that independent of dimension that this should be true, okay, and again I don't understand the mathematics behind this, but what this, one thing that, that this does say is, so, so this gives us what I'll, the last theorem, theorem three, if RMK is again say complete, and dimension four, and F regular, so not, not just weakly F regular, F regular, so I do need to know that when I localize I'm still, still weakly F regular, then R is strongly F regular, and this is because once we know that we're F, weakly F regular in dimension three, we are strongly F regular, and then the, the anti-canonical cover is noetherian on the punctured spectrum, and now we have the conditions of theorem two, satisfied. I'll stop there. Any questions for Yen? I have an elementary question, what is the meaning of principle and co-dimension two? Just that it's generated by, by a single element, it's a principle ideal at all height two, at height two primes, at height two primes. It can be different generators, that's right, I mean essentially what happens is, yeah, you, you kind of get a, that, that generator A works everywhere, but finitely many primes, and then you have to find another generator at a few primes, or of course it might be a higher power, sorry, it's a power of A works. Any other questions, comments? So, I have a comment, this reminds me a lot of, was somewhat of your work with Janet on uniform, art and reese, seems to also involve a lot of painful estimation, are there any common, common themes in this? You're going through a lot of like Kozoo, you know, analysis of Kozoo homology and, you know, you're looking at. So, the thing about the, the, the uniform art and reese type theorem is, is that because you can get to reductions, you, you I think end up kind of needing to annihilate parameter homology, and that's, you know, easy in some sense, right, because, because we, we know we have those, and here you're, you're having to annihilate the, the homology or co-homology of, of something involving one extra, you know, it's, it's, it's an almost complete intersection of smaller size than the, the ring, and that, that's where the problem lies. So essentially you, you, you have, we, we got rid of x1, so we're really looking at x2 through xd, taking the q-th powers, but we also have a in there, that, and a shares primes with x2, that throws everything off. You secretly love this kind of pain, don't you? It's, I, I'd have to argue it's not so secret at this point. Any other questions? All right, let's sing Yannick.